7.7 The Inverse Hyperbolic Functions
| Calculus Of One Real Variable – By Pheng Kim Ving |
| 7.7 |
Return To Contents
Go To Problems & Solutions
The Inverse Hyperbolic Sine Function
The graph of the hyperbolic sine function y = sinh x is sketched in Fig. 1.1. Clearly sinh is one-to-one, and so has an
inverse, denoted sinh–1. The inverse hyperbolic sine function sinh–1 is defined as follows:
The graph of y = sinh–1 x is the mirror image of that of y = sinh x in the line y = x. It’s shown in Fig. 1.1. We have
dom(sinh–1) = R and range(sinh–1) = R.
| Fig. 1.1 Graph of y = sinh–1 x. |
The Inverse Hyperbolic Cosine Function
| Fig. 1.2 Graph of y = cosh–1 x. |
The Inverse Hyperbolic Tangent Function
The graph of the hyperbolic tangent function y = tanh x is sketched in Fig. 1.3. Clearly tanh is one-to-one, and so has an
inverse, denoted tanh–1. The inverse hyperbolic tangent function tanh–1 is defined as follows:
| Fig. 1.3 Graph of y = tanh–1 x. |
The Inverse Hyperbolic Cotangent Function
The graph of the hyperbolic cotangent function y = coth x is sketched in Fig. 1.4. Clearly coth is one-to-one, and thus has
an inverse, denoted coth–1. The inverse hyperbolic cotangent function coth–1 is defined as follows:
| Fig. 1.4 Graph of y = coth–1 x. |
The Inverse Hyperbolic Secant Function
| Fig. 1.5 Graph of y = sech–1 x. |
The Inverse Hyperbolic Cosecant Function
The graph of the hyperbolic cosecant function y = csch x is sketched in Fig. 1.6. Clearly csch is one-to-one, and so has
an inverse, denoted csch–1. The inverse hyperbolic cosecant function csch–1 is defined as follows:
| Fig. 1.6 Graph of y = csch–1 x. |
Example 1.1
Prove the identity:
Note
Recall that the inverse of the natural exponential function is the natural logarithm function. Since the hyperbolic functions are defined in terms of the natural exponential function, it’s not surprising that their inverses can be expressed in terms
of the natural logarithm function. Also see Problem & Solution 1 and Problem & Solution 2.
Solution
Let y = sinh–1 x. Then x = sinh y = (ey – e–y)/2. So ey – e–y – 2x = 0. Multiplying both sides by ey yields e2y – 1 – 2xey = 0,
or e2y – 2xey – 1 = 0, which is a quadratic equation in ey. Its roots are:
EOS
Go To Problems & Solutions Return To Top Of Page
We prove formula [2.1] as follows. Let y = sinh–1 x. Then x = sinh y. Differentiating this equation implicitly with respect to x
we get:
The remaining differentiation formulas are proved in a similar way.
Example 2.1
Differentiate sinh–1 tan x.
Solution
EOS
Return To Top Of Page
1. In Example 1.1 we proved the identity:
Also see Problem & Solution 2.
Solution
Let y = cosh–1 x. Then x = cosh y = (ey + e–y)/2. So ey + e–y – 2x = 0. Multiplying both sides by ey yields e2y + 1 – 2xey =
0, or e2y – 2xey + 1 = 0, which is a quadratic equation in ey. Its roots are:
Return To Top Of Page
2. In Example 1.1 we proved 1 identity and in Problem & Solution 1 you were asked to prove another identity. Now again
you’re asked to prove the following 2 identities:
Solution
a. Let y = tanh–1 x. So x = tanh y and |x| < 1. We have:
xe2y – x = e2y + 1,
e2y(x – 1) = x + 1,
Return To Top Of Page
3. Differentiate the following functions.
a. sinh–1 (x/a), a > 0.
b. cosh–1 (x/a), a > 0.
Solution
Return To Top Of Page
4. Differentiate the following functions.
a. y = sech–1 (x2).
b. f(t) = csch–1 tan t.
Solution
Return To Top Of Page
5. Prove that:
Solution
Let y = csch–1 x. Then x = csch y. Let z = sinh–1 (1/x), so that 1/x = sinh z, or:
Return To Top Of Page Return To Contents